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A system consists of three particles located at the corners of a right triangle as shown in the figure. Fin the position vector of centre of mass of the system. |
Answer» Solution : Using the equation `X_(c )=(summ_(i)x_(i))/(M)=(2md+m(B+d)+3m(d+b))/(6M)=d+((2)/(3))b` `Y_(c )=(Summ_(i)y_(i))/(M)=(2m(0)+m(0)+3mh)/(6m)=(H)/(2)` `Z_(c )=0`, because the particles are in X-Y plane we can express the position of centre of mass from the origin using a position vector as `vecr_(c )=X_(c )hati+Y_(c )hatj+Z_(c )hatk,vecr_(c )=(d+(2)/(3)b)hati+(h)/(2)hatj` |
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